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Veterans Upward Bound Algebra I Concepts - Honors Brenda Meery Kaitlyn Spong Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org C HAPTER 4 Exponents Chapter Outline 4.1 Z ERO AND N EGATIVE E XPONENTS 4.2 P RODUCT R ULES FOR E XPONENTS 4.3 Q UOTIENT R ULES FOR E XPONENTS 4.4 P OWER R ULE FOR E XPONENTS 4.5 E XPONENTIAL E XPRESSIONS 4.6 S CIENTIFIC N OTATION Introduction Here you’ll learn all about exponents in algebra. You will learn the properties of exponents and how to simplify exponential expressions. You will learn how exponents can help you write very large or very small numbers with scientific notation. You will also learn how to solve different types of exponential equations where the variable appears as the exponent or the base. Finally, you will explore different types of exponential functions of the form x x y = bx , y = abx , y = ab c , and y = ab c + d as well as applications of exponential functions. 314 www.ck12.org Chapter 4. Exponents 4.1 Zero and Negative Exponents Here you’ll learn how to work with zero and negative exponents. How can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent? Watch This Khan Academy Negative Exponent Intuition MEDIA Click image to the left for more content. Guidance Zero Exponent Recall that am = am−n an . If m = n, then the following would be true: am = am−n = a0 an 33 = 33−3 = 30 33 However, any quantity divided by itself is equal to one. Therefore, general: 33 33 = 1 which means 30 = 1. This is true in a0 = 1 if a 6= 0. Note that if a = 0, 00 is not defined. Negative Exponents 42 × 4−2 = 42+(−2) = 40 = 1 Therefore: 315 4.1. Zero and Negative Exponents www.ck12.org 42 × 4−2 = 1 42 × 4−2 1 = 2 2 4 4 2 −2 4 ×4 1 = 2 2 4 4 4−2 = Divide both sides by 42 . Simplify the equation. 1 42 This is true in general and creates the following laws for negative exponents: • a−m = 1 am • 1 a−m = am These laws for negative exponents can be expressed in many ways: • If a term has a negative exponent, write it as 1 over the term with a positive exponent. For example: a−m = a1m 1 and a−m = am −2 2 • If a term has a negative exponent, write the reciprocal with a positive exponent. For example: 23 = 32 −m and a−m = a 1 = a1m • If the term is a factor in the numerator with a negative exponent, write it in the denominator with a positive n exponent. For example: 3x−3 y = 3y and a−m bn = a1m (bn ) = abm x3 • If the term is a factor in the denominator with a negative exponent, write it in the numerator with a positive 3 n m exponent. For example: x2x−2 = 2x3 (x2 ) and ab−m = bn a1 = bn am These ways for understanding negative exponents provide shortcuts for arriving at solutions without doing tedious calculations. The results will be the same. Example A Evaluate the following using the laws of exponents. 3 −2 4 Solution: There are two methods that can be used to evaluate the expression. Method 1: Apply the negative exponent rule a−m = 316 1 am www.ck12.org Chapter 4. Exponents −2 3 = 4 1 3 2 Write the expression with a positive exponent by applying a−m = 4 1 1 = 32 2 3 42 4 1 = 32 42 1 1 a n b = an bn = an bn Evaluate the powers. 9 16 = 1÷ 9 16 Apply the law of exponents for raising a quotient to a power. 1 . am 9 16 Divide 9 16 16 = 1× = 16 9 9 −2 3 16 = 4 9 1÷ Method 2: Apply the shortcut and write the reciprocal with a positive exponent. −2 2 3 4 = 4 3 2 4 42 = 2 3 3 42 16 = 32 9 −2 3 16 = 4 9 Write the reciprocal with a positive exponent. Apply the law of exponents for raising a quotient to a power. a n b Simplify. Applying the shortcut facilitates the process for obtaining the solution. Example B State the following using only positive exponents: (If possible, use shortcuts) i) y−6 −3 ii) ab iii) x5 y−4 iv) a2 × a−5 Solutions: i) y−6 y−6 = Write the expression with a positive exponent by applying a−m = 1 . am 1 y6 317 4.1. Zero and Negative Exponents www.ck12.org ii) a −3 Write the reciprocal with a positive exponent. b a −3 b 3 = b a 3 b b3 = 3 a a a −3 b3 = 3 b a a n Apply the law of exponents for raising a quotient to a power. b = an bn iii) x5 y−4 Apply the negative exponent rule. x5 = x5 y−4 y4 1 1 a−m = am Simplify. x5 = x 5 y4 y−4 iv) a2 × a−5 Apply the product rule for exponents am × an = am+n . a2 × a−5 = a2+(−5) Simplify. a2+(−5) = a−3 Write the expression with a positive exponent by applying a−m = a−3 = 1 a3 a2 × a−5 = 1 a3 Example C Evaluate the following: 7−2 +7−1 7−3 +7−4 Solution: There are two methods that can be used to evaluate the problem. Method 1: Work with the terms in the problem in exponential form. Numerator: 318 1 . am www.ck12.org Chapter 4. Exponents 7−2 = 1 1 and 7−1 = 2 7 7 Apply the definition a−m = 1 1 + 72 7 1 1 7 + 72 7 7 7 1+7 8 1 + 2= 2 = 2 2 7 7 7 7 1 am A common denominator is needed to add the fractions. Multiply 1 7 by to obtain the common denominator of 72 7 7 Add the fractions. Denominator: 1 1 and 7−4 = 4 3 7 7 1 1 + 73 74 7 1 1 + 4 3 7 7 7 1 1+7 8 7 + 4= 4 = 4 4 7 7 7 7 7−3 = Apply the definition a−m = 1 am A common denominator is needed to add the fractions. Multiply 1 7 by to obtain the common denominator of 74 3 7 7 Add the fractions. Numerator and Denominator: 8 8 ÷ 4 2 7 7 8 74 × 72 8 74 74 8 × = 2 = 72 = 49 72 7 8 Divide the numerator by the denominator. Multiply by the reciprocal. Simplify. 7−2 + 7−1 = 49 7−3 + 7−4 Method 2: Multiply the numerator and the denominator by 74 . This will change all negative exponents to positive exponents. Apply the product rule for exponents and work with the terms in exponential form. 7−2 + 7−1 7−3 + 7−4 4 −2 7 7 + 7−1 74 7−3 + 7−4 72 + 73 71 + 70 49 + 343 392 = = 49 7+1 8 Apply the distributive property with the product rule for exponents. Evaluate the numerator and the denominator. 7−2 + 7−1 = 49 7−3 + 7−4 Whichever method is used, the result is the same. 319 4.1. Zero and Negative Exponents www.ck12.org Concept Problem Revisited m By the quotient rule for exponents, xxm = xm−m = x0 . Since anything divided by itself is equal to 1 (besides 0), Therefore, x0 = 1 as long as x 6= 0. Also by the quotient rule for exponents, you would have x2 x5 = x·x x·x·x·x·x = 1 . x3 x2 x5 xm xm = 1. = x2−5 = x−3 . If you were to expand and reduce the original expression Therefore, x−3 = 1 . x3 This generalizes to x−a = 1 xa . Vocabulary Base In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base. Exponent In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are: In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 × 2 × 2 × 2. In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here: (−3y)4 = −3y × −3y × −3y × −3y. Laws of Exponents The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions. Guided Practice 1. Use the laws of exponents to simplify the following: (−3x2 )3 (9x4 y)−2 2. Rewrite the following using only positive exponents. (x2 y−1 )2 3. Use the laws of exponents to evaluate the following: [5−4 × (25)3 ]2 Answers: 1. 320 www.ck12.org Chapter 4. Exponents (−3x2 )3 (9x4 y)−2 Apply the laws of exponents (am )n = amn and a−m = 1 am 1 (−3x ) (9x y) = (−3 x ) Simplify and apply (ab)n = an bn (9x4 y)2 1 1 6 (−33 x6 ) = −27x Simplify. (9x4 y)2 (92 x8 y2 ) am −27x6 1 6 Simplify and apply the quotient rule for exponents = am−n . = − 27x (92 x8 y2 ) 81x8 y2 an 2 3 4 −2 3 6 1x−2 −27x6 = − 81x8 y2 3y2 Apply the negative exponent rule a−m = (−3x2 )3 (9x4 y)−2 = − 1 am 1 3x2 y2 2. (x2 y−1 )2 = x4 y−2 = x4 y2 3. [5−4 × (25)3 ]2 Try to do this one by applying the laws of exponents. [5 −4 3 2 × (25) ] = [5 −4 × (5 ) ] [5 −4 2 3 2 −4 × 56 ]2 × (5 ) ] = [5 2 3 2 [5−4 × 56 ]2 = (52 )2 (52 )2 = 54 54 = 625 [5−4 × (25)3 ]2 = 54 = 625 Practice Evaluate each of the following expressions: 2 0 3 −2 − 25 (−3)−3 1. − 2. 3. −2 4. 6 × 12 5. 7−4 × 74 Rewrite the following using positive exponents only. Simplify where possible. 6. (4wx−2 y3 z−4 )3 321 4.1. Zero and Negative Exponents 2 3 −2 7. da−2bbcc −6 8. x−2 (x2 − 1) 9. m4 (m2 + m − 5m−2 ) −2 −2 10. xx−1 yy−1 −2 3 −4 −7 y 11. xy4 x6 12. 13. 14. 15. 322 (x−2 y4 )2 (x5 y−3 )4 (3xy2 )3 (3x2 y)4 0 x2 y−25 z5 −12.4x3 y −2 5 −2 −3 y x y3 x4 www.ck12.org www.ck12.org Chapter 4. Exponents 4.2 Product Rules for Exponents Here you’ll learn how to multiply two terms with the same base and how to find the power of a product. Suppose you have the expression: x·x·x·x·x·x·x·x·x·y·y·y·y·y·x·x·x·x How could you write this expression in a more concise way? Watch This MEDIA Click image to the left for more content. James Sousa: Exponential Notation Guidance In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself. For example: • x3 = x · x · x • 24 = 2 · 2 · 2 · 2 = 16. There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know so that you can work with expressions and equations that involve exponents more easily. Here you will learn two rules that have to do with exponents and products. RULE: To multiply two terms with the same base, add the exponents. am × an = (a × a × . . . × a) (a × a × . . . × a) ←−−−−−−−−−→ ←−−−−−−−−−→ ↓ ↓ m factors m n m n n factors a × a = (a × a × a . . . × a) ←−−−−−−−−−−→ ↓ m + n factors a ×a = a m+n 323 4.2. Product Rules for Exponents www.ck12.org RULE: To raise a product to a power, raise each of the factors to the power. (ab)n = (ab) × (ab) × . . . × (ab) ←−−−−−−−−−−−−−−→ ↓ n factors n (ab) = (a × a × . . . × a) × (b × b × . . . × b) ←−−−−−−−−−→ ←−−−−−−−−−→ ↓ ↓ n factors n n factors n n (ab) = a b Example A Simplify 32 × 33 . Solution: 32 × 33 The base is 3. 32+3 Keep the base of 3 and add the exponents. 3 5 This answer is in exponential form. The answer can be taken one step further. The base is numerical so the term can be evaluated. 35 = 3 × 3 × 3 × 3 × 3 35 = 243 32 × 33 = 35 = 243 Example B Simplify (x3 )(x6 ). Solution: (x3 )(x6 ) x 3+6 x9 (x3 )(x6 ) = x9 Example C Simplify y5 · y2 . 324 The base is x. Keep the base of x and add the exponents. The answer is in exponential form. www.ck12.org Chapter 4. Exponents Solution: y5 · y2 5+2 The base is y. y Keep the base of y and add the exponents. y7 The answer is in exponential form. y5 · y2 = y7 Example D Simplify 5x2 y3 · 3xy2 . Solution: 5x2 y3 · 3xy2 The bases are x and y. 15(x2 y3 )(xy2 ) Multiply the coefficients - 5 × 3 = 15. Keep the base of x and y and add the exponents of the same base. If a base does not have a written exponent, it is understood as 1. 15x 2+1 3+2 y 15x3 y5 The answer is in exponential form. 5x2 y3 · 3xy2 = 15x3 y5 Concept Problem Revisited x · x · x · x · x · x · x · x · x · y · y · y · y · y · x · x · x · x can be rewritten as x9 y5 x4 . Then, you can use the rules of exponents to simplify the expression to x13 y5 . This is certainly much quicker to write! Vocabulary Base In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base. Exponent In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are: In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 × 2 × 2 × 2. In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here: (−3y)4 = −3y × −3y × −3y × −3y. Laws of Exponents The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions. 325 4.2. Product Rules for Exponents www.ck12.org Guided Practice Simplify each of the following expressions. 1. (−3x)2 2. (5xy)3 3. (23 · 32 )2 Answers: 1. 9x2 . Here are the steps: (−3x)2 = (−3)2 · (x)2 = 9x2 2. 125x3 y3 . Here are the steps: (5x2 y4 )3 = (5)3 · (x)3 · (y)3 = 125x3 y3 3. 5184. Here are the steps: (23 · 32 )2 = (8 · 9)2 = (72)2 = 5184 OR (23 · 32 )2 = (8 · 9)2 = 82 · 92 = 64 · 81 = 5184 Practice Simplify each of the following expressions, if possible. 1. 2. 3. 4. 5. 6. 7. 8. 326 42 × 44 x4 · x12 (3x2 y4 )(9xy5 z) (2xy)2 (4x2 y3 ) (3x)5 (2x)2 (3x4 ) x3 y2 z · 4xy2 z7 x2 y3 + xy2 (0.1xy)4 www.ck12.org 9. 10. 11. 12. Chapter 4. Exponents (xyz)6 2x4 (x2 − y2 ) 3x5 − x2 3x8 (x2 − y4 ) Expand and then simplify each of the following expressions. 13. (x5 )3 14. (x6 )8 15. (xa )b Hint: Look for a pattern in the previous two problems. 327 4.3. Quotient Rules for Exponents www.ck12.org 4.3 Quotient Rules for Exponents Here you’ll learn how to divide two terms with the same base and find the power of a quotient. Suppose you have the expression: x·x·x·x·x·x·x·x·x·y·y·y·y·y x·x·x·x·x·x·y·y·y How could you write this expression in a more concise way? Watch This MEDIA Click image to the left for more content. James Sousa: Simplify Exponential Expressions- Quotient Rule Guidance In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself. For example: • x3 = x · x · x • 24 = 2 · 2 · 2 · 2 = 16. There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know so that you can work with expressions and equations that involve exponents more easily. Here you will learn two rules that have to do with exponents and quotients. RULE: To divide two powers with the same base, subtract the exponents. 328 www.ck12.org Chapter 4. Exponents m factors ↑ ← − − − − − −−−−→ am (a × a × . . . × a) = m > n; a 6= 0 an (a × a × . . . × a) ←−−−−−−−−−→ ↓ n factors am an = (a × a × . . . × a) ←−−−−−−−−−→ ↓ m − n factors am an = am−n RULE: To raise a quotient to a power, raise both the numerator and the denominator to the power. a n b = a a a × ×...× b b b ←−−−−−−−−−→ ↓ n factors n factors ↑ ← − − − − − − −→ a n (a × a × . .−.− × a) = b (b × b × . . . × b) ←−−−−−−−−−→ ↓ a n b n factors an = n (b 6= 0) b Example A Simplify 27 ÷ 23 . Solution: 27 ÷ 23 The base is 2. 2 7−3 Keep the base of 2 and subtract the exponents. 2 4 The answer is in exponential form. The answer can be taken one step further. The base is numerical so the term can be evaluated. 329 4.3. Quotient Rules for Exponents www.ck12.org 24 = 2 × 2 × 2 × 2 24 = 16 27 ÷ 23 = 24 = 16 Example B Simplify x8 . x2 Solution: x8 x2 x8−2 x The base is x. Keep the base of x and subtract the exponents. 6 The answer is in exponential form. x8 = x6 x2 Example C Simplify 16x5 y5 . 4x2 y3 Solution: 16x5 y5 4x2 y3 5 5 x y 4 2 3 x y The bases are x and y. Divide the coefficients - 16 ÷ 4 = 4. Keep the base of x and y and subtract the exponents of the same base. 4x5−2 y5−3 4x3 y2 Concept Problem Revisited x·x·x·x·x·x·x·x·x·y·y·y·y·y x·x·x·x·x·x·y·y·y can be rewritten as x9 y5 x6 y3 and then simplified to x3 y2 . Vocabulary Base In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base. Exponent In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are: 330 www.ck12.org Chapter 4. Exponents In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 × 2 × 2 × 2. In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here: (−3y)4 = −3y × −3y × −3y × −3y. Laws of Exponents The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions. Guided Practice Simplify each of the following expressions. 2 1. 32 3 2. 6x 2 3. 3x 4y Answers: 2 2 1. 23 = 232 = 94 3 3 x3 2. 6x = 6x3 = 216 2 2 2 9x2 3. 3x = 342 xy2 = 16y 2 4y Practice Simplify each of the following expressions, if possible. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 2 6 5 4 3 7 4 x y 20x4 y5 5x2 y4 42x2 y8 z2 4 6xyz3 3x 4y 72x2 y4 8x2 y3 x 5 4 24x14 y8 3x5 y7 72x3 y9 6 24xy3 7 y 20x12 −5x8 13. Simplify using the laws of exponents: 23 25 14. Evaluate the numerator and denominator separately and then simplify the fraction: 15. Use your result from the previous problem to determine the value of a: 23 25 = 23 25 1 2a 331 4.3. Quotient Rules for Exponents 16. Use your results from the previous three problems to help you evaluate 2−4 . 332 www.ck12.org www.ck12.org Chapter 4. Exponents 4.4 Power Rule for Exponents Here you’ll learn how to find the power of a power. Can you simplify an expression where an exponent has an exponent? For example, how would you simplify [(23 )2 ]4 ? Watch This MEDIA Click image to the left for more content. James Sousa: Properties of Exponents Guidance In the expression x3 , the x is called the base and the 3 is called the exponent. Exponents are often referred to as powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself. For example: • x3 = x · x · x • 24 = 2 · 2 · 2 · 2 = 16. There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know so that you can work with expressions and equations that involve exponents more easily. Here you will learn a rule that has to do with raising a power to another power. RULE: To raise a power to a new power, multiply the exponents. (am )n = (a × a × . . . × a)n ←−−−−−−−−−−→ ↓ m factors m n (a ) = (a × a × . . . × a) × (a × a × . . . × a) (a × a × . . . × a) ←−−−−−−−−−→ ←−−−−−−−−−→ ←−−−−−−−−−→ ↓ ↓ ↓ m factors m factors m factors ←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ n times (am )n = a × a × a . . . × a ←−−−−−−−−−→ mn factors (am )n = amn 333 4.4. Power Rule for Exponents www.ck12.org Example A Evaluate (23 )2 . Solution: (23 )2 = 26 = 64. Example B Simplify (x7 )4 . Solution: (x7 )4 = x28 . Example C Evaluate (32 )3 . Solution: (32 )3 = 36 = 729. Example D Simplify (x2 y4 )2 · (xy4 )3 . Solution: (x2 y4 )2 · (xy4 )3 = x4 y8 · x3 y12 = x7 y20 . Concept Problem Revisited [(23 )2 ]4 = [26 ]4 = 224 . Notice that the power rule applies even when a number has been raised to more than one power. The overall exponent is 24 which is 3 · 2 · 4. Vocabulary Base In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base. Exponent In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are: In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 × 2 × 2 × 2. In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here: (−3y)4 = −3y × −3y × −3y × −3y. Laws of Exponents The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions. 334 www.ck12.org Chapter 4. Exponents Guided Practice You know you can rewrite 24 as 2 × 2 × 2 × 2 and then calculate in order to find that 24 = 16 . This concept can also be reversed. To write 32 as a power of 2, 32 = 2 × 2 × 2 × 2 × 2. There are 5 twos; therefore, 32 = 25 . Use this idea to complete the following problems. 1. Write 81 as a power of 3. 2. Write (9)3 as a power of 3. 3. Write (43 )2 as a power of 2. Answers: 1. 81 = 3 × 3 = 9 × 3 = 27 × 3 = 81 There are 4 threes. Therefore 81 = 34 2. 9 = 3 × 3 = 9 There are 2 threes. Therefore 9 = 32 . (32 )3 Apply the law of exponents for power to a power-multiply the exponents. 32×3 = 36 Therefore (9)3 = 36 3. 4 = 2 × 2 = 4 There are 2 twos. Therefore 4 = 22 (22 )3 2 Apply the law of exponents for power to a power-multiply the exponents. 22×3 = 26 (26 )2 Apply the law of exponents for power to a power-multiply the exponents. 26×2 = 212 Therefore (43 )2 = 212 335 4.4. Power Rule for Exponents Practice Simplify each of the following expressions. 1. 2. 3. 4 5 x y3 (5x2 y4 )5 (5xy2 )3 x 8 y9 (x2 y)3 (x2 y4 )3 4. 5. (3x2 )2 · (4xy4 )2 6. (2x3 y5 )(5x2 y)3 7. (x4 y6 z2 )2 (3xyz)3 2 4 x 8. 2y 3 9. 10. 11. 12. 13. 14. 15. 16. 336 (4xy3 )4 (2xy2 )3 True or false: (x2 + y3 )2 = x4 + y6 True or false: (x2 y3 )2 = x4 y6 Write 64 as a power of 4. Write (16)3 as a power of 2. Write (94 )2 as a power of 3. Write (81)2 as a power of 3. Write (253 )4 as a power of 5. www.ck12.org www.ck12.org Chapter 4. Exponents 4.5 Exponential Expressions Here you’ll learn how to use all of the laws of exponents to simplify and evaluate exponential expressions. Can you simplify the following expression so that it has only positive exponents? 8x3 y−2 (−4a2 b4 )−2 Watch This James Sousa: Simplify Exponential Expressions MEDIA Click image to the left for more content. Guidance The following table summarizes all of the rules for exponents. Laws of Exponents If a ∈ R, a ≥ 0 and m, n ∈ Q, then 1. 2. 3. 4. 5. 6. 7. 8. am × an = am+n am m−n (if m > n, a 6= 0) an = a m n (a ) = amn n n n (ab) = aan b a n = bn (b 6= 0) b a0 = 1 (a 6= 0) a−m = a1m √ √ m m n a n = am = n a Example A 1 Evaluate 81− 4 . Solution: First, rewrite with a positive exponent: 1 1 1 4 81− 4 = 11 = 81 . 81 4 Next, evaluate the fractional exponent: r 1 1 4 1 4 = =1 81 81 3 337 4.5. Exponential Expressions www.ck12.org Example B Simplify (4x3 y)(3x5 y2 )4 . Solution: (4x3 y)(3x5 y2 )4 = (4x3 y)(81x20 y8 ) = 324x23 y9 Example C Simplify x−2 y x 4 y3 −2 . Solution: x−2 y x 4 y3 −2 = x 4 y3 x−2 y 2 = (x6 y2 )2 = x12 y4 Concept Problem Revisited 8x3 y−2 = (8x3 y−2 )(−4x2 y4 )2 (−4x2 y4 )−2 = (8x3 y−2 )(16x4 y8 ) = 8 · 16 · x3 · x4 · y−2 · y8 = 128x7 y6 Vocabulary Base In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base. Exponent In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are: In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 × 2 × 2 × 2. In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here: (−3y)4 = −3y × −3y × −3y × −3y. Laws of Exponents The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions. 338 www.ck12.org Chapter 4. Exponents Guided Practice Use the laws of exponents to simplify each of the following: 1. (−2x)5 (2x2 ) 2. (16x10 ) 43 x5 3. (x15 )(x24 )(x25 ) (x7 )8 Answers: 1. (−2x)5 (2x2 ) = (−32x5 )(2x2 ) = −64x7 2. (16x10 ) 43 x5 = 12x15 3. (x15 )(x24 )(x25 ) (x7 )8 = x64 x56 = x8 Practice Simplify each expression. 1. (x10 )(x10 ) 2. (7x3 )(3x7 ) 3. (x3 y2 )(xy3 )(x5 y) 3 2) 4. (x(x)(x 4) 5. x2 x−3 x 6 y8 x4 y−2 (2x12 )3 6. 7. 8. (x5 y10 )7 10 3 9. 2x 3y20 Express each of the following as a power of 3. Do not evaluate. 10. 11. 12. 13. 14. (33 )5 (39 )(33 ) (9)(37 ) 94 (9)(272 ) Apply the laws of exponents to evaluate each of the following without using a calculator. 15. 16. 17. 18. 19. (23 )(22 ) 66 ÷ 65 −(32 )3 (12 )3 + (13 )2 1 6 1 8 ÷ 3 3 Use the laws of exponents to simplify each of the following. 20. (4x)2 21. (−3x)3 339 4.5. Exponential Expressions 22. 23. 24. 25. 26. (x3 )4 (3x)(x7 ) (5x)(4x4 ) (−3x2 )(−6x3 ) (10x8 ) ÷ (2x4 ) Simplify each of the following using the laws of exponents. 1 1 27. 5 2 × 5 3 4 8 12 14 28. (d se f ) 1 √ xy 4 y2 29. 2 x3 1 30. (32a20 b−15 ) 5 2 31. (729x12 y−6 ) 3 340 www.ck12.org www.ck12.org Chapter 4. Exponents 4.6 Scientific Notation Here you’ll learn about scientific notation. Very large and very small quantities and measures are often used to provide information in magazines, textbooks, television, newspapers and on the Internet. Some examples are: • The distance between the sun and Neptune is 4,500,000,000 km. • The diameter of an electron is approximately 0.00000000000022 inches. Scientific notation is a convenient way to represent such numbers. How could you write the numbers above using scientific notation? Watch This Khan Academy Scientific Notation MEDIA Click image to the left for more content. Khan Academy Scientific Notation Examples MEDIA Click image to the left for more content. Guidance To represent a number in scientific notation means to express the number as a product of two factors: a number between 1 and 10 (including 1) and a power of 10. A positive real number ’x’ is said to be written in scientific notation if it is expressed as x = a × 10n where 1 ≤ a < 10 and n ∈ Z. In other words, a number in scientific notation is a single nonzero digit followed by a decimal point and other digits, all multiplied by a power of 10. 341 4.6. Scientific Notation www.ck12.org When working with numbers written in scientific notation, you can use the following rules. These rules are proved by example in Example B and Example C. (A × 10n ) + (B × 10n ) = (A + B) × 10n (A × 10n ) − (B × 10n ) = (A − B) × 10n (A × 10m ) × (B × 10n ) = (A × B) × (10m+n ) (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n ) Example A Write the following numbers using scientific notation: i) 2,679,000 ii) 0.00005728 Solutions: i) 2, 679, 000 = 2.679 × 1, 000, 000 2.679 × 1, 000, 000 = 2.679 × 106 The exponent, n = 6, represents the decimal point that is 6 places to the right of the standard position of the decimal point. ii) 0.00005728 = 5.728 × 0.00001 1 5.728 × 0.00001 = 5.728 × 100, 000 1 1 5.728 × = 5.728 × 5 100, 000 10 1 5.728 × = 5.728 × 10−5 100, 000 The exponent, n = −5, represents the decimal point that is 5 places to the left of the standard position of the decimal point. One advantage of scientific notation is that calculations with large or small numbers can be done by applying the laws of exponents. 342 www.ck12.org Chapter 4. Exponents Example B Complete the following table. 343 4.6. Scientific Notation www.ck12.org TABLE 4.1: Expression in Scientific Notation 1.3 × 105 + 2.5 × 105 3.7 × 10−2 + 5.1 × 10−2 4.6 × 104 − 2.2 × 104 7.9 × 10−2 − 5.4 × 10−2 Expression in Standard Form Result Form in Standard Result in Scientific Notation Result in Form 380,000 0.088 24,000 0.025 Standard Result in Scientific Notation 3.8 × 105 8.8 × 10−2 2.4 × 104 2.5 × 10−2 Solution: TABLE 4.2: Expression in Scientific Notation 1.3 × 105 + 2.5 × 105 3.7 × 10−2 + 5.1 × 10−2 4.6 × 104 − 2.2 × 104 7.9 × 10−2 − 5.4 × 10−2 Expression in Standard Form 130, 000 + 250, 000 0.037 + 0.051 46, 000 − 22, 000 0.079 − 0.054 Note that the numbers in the last column have the same power of 10 as those in the first column. Example C Complete the following table. TABLE 4.3: Expression in Scientific Notation (3.6 × 102 ) × (1.4 × 103 ) (2.5 × 103 ) × (1.1 × 10−6 ) (4.4 × 104 ) ÷ (2.2 × 102 ) (6.8 × 10−4 ) ÷ (3.2 × 10−2 ) Expression in Standard Form Result Form in Standard Result in Scientific Notation Result in Form 504,000 0.00275 200 0.02125 Standard Result in Scientific Notation 5.04 × 105 2.75 × 10−3 2.0 × 102 2.125 × 10−2 Solution: TABLE 4.4: Expression in Scientific Notation (3.6 × 102 ) × (1.4 × 103 ) (2.5 × 103 ) × (1.1 × 10−6 ) (4.4 × 104 ) ÷ (2.2 × 102 ) (6.8 × 10−4 ) ÷ (3.2 × 10−2 ) Expression in Standard Form 360 × 1400 2500 × 0.0000011 44, 000 ÷ 220 0.00068 ÷ 0.032 Note that for multiplication, the power of 10 is the result of adding the exponents of the powers in the first column. For division, the power of 10 is the result of subtracting the exponents of the powers in the first column. 344 www.ck12.org Chapter 4. Exponents Example D Calculate each of the following: i) 4.6 × 104 + 5.3 × 105 ii) 4.7 × 10−3 − 2.4 × 10−4 iii) (7.3 × 105 ) × (6.8 × 104 ) iv) (4.8 × 109 ) ÷ (5.79 × 107 ) Solution: i) Before the rule (A × 10n ) + (B × 10n ) = (A + B) × 10n can be used, one of the numbers must be rewritten so that the powers of 10 are the same. Rewrite 4.6 × 104 4.6 × 104 = (0.46 × 101 ) × 104 The power 101 indicates the number of places to the right that the decimal point must be moved to return 0.46 to the original number of 4.6. (0.46 × 101 ) × 104 = 0.46 × 105 Add the exponents of the power. Rewrite the question and substitute 4.6 × 104 with 0.46 × 105 . 0.46 × 105 + 5.3 × 105 Apply the rule (A × 10n ) + (B × 10n ) = (A + B) × 10n . (0.46 × 105 ) + (5.3 × 105 ) = (0.46 + 5.3) × 105 (0.46 + 5.3) × 105 = 5.76 × 105 4.6 × 104 + 5.3 × 105 = 5.76 × 105 ii) Before the rule (A × 10n ) − (B × 10n ) = (A − B) × 10n can be used, one of the numbers must be rewritten so that the powers of 10 are the same. Rewrite 4.7 × 10−3 4.7 × 10−3 = (47 × 10−1 ) × 10−3 The power 10−1 indicates the number of places to the left that the decimal point must be moved to return 47 to the original number of 4.7. (47 × 10−1 ) × 10−3 = 47 × 10−4 Add the exponents of the power. Rewrite the question and substitute 4.7 × 10−3 with 47 × 10−4 . 47 × 10−4 − 2.4 × 10−4 Apply the rule (A × 10n ) − (B × 10n ) = (A − B) × 10n 345 4.6. Scientific Notation www.ck12.org . (47 × 10−4 ) − (2.4 × 10−4 ) = (47 − 2.4) × 10−4 (47 × 10−4 ) − (2.4 × 10−4 ) = 44.6 × 10−4 The answer must be written in scientific notation. 44.6 × 10−4 = (4.46 × 101 ) × 10−4 −4 4.46 × 10 × 10 Apply the law of exponents − add the exponents of the power. −3 = 4.46 × 10 4.7 × 10−3 − 2.4 × 10−4 = 4.46 × 10−3 iii) (7.3 × 105 ) × (6.8 × 104 ) 7.3 × 105 × 6.8 × 104 Apply the rule (A × 10m ) × (B × 10n ) = (A × B) × (10m+n ) . (7.3 × 105 ) × (6.8 × 104 ) = (7.3 × 6.8) × (105+4 ) (7.3 × 6.8) × (105+4 ) = (49.64) × (109 ) (49.64) × (109 ) = 49.64 × 109 9 1 Write the answer in scientific notation. 9 49.64 × 10 = (4.964 × 10 ) × 10 Apply the law of exponents − add the exponents of the power. 49.64 × 109 = 4.964 × 1010 (7.3 × 105 ) × (6.8 × 104 ) = 4.964 × 1010 iv) (4.8 × 109 ) ÷ (5.79 × 107 ) (4.8 × 109 ) ÷ (5.79 × 107 ) Apply the rule (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n ) . (4.8 × 109 ) ÷ (5.79 × 107 ) = (4.8 ÷ 5.79) × 109−7 Apply the law of exponents − subtract the exponents of the power. (4.8 ÷ 5.79) × 109−7 = (0.829) × 102 2 −1 2 (0.829) × 10 = (8.29 × 10 ) × 10 Write the answer in scientific notation. Apply the law of exponents − add the exponents of the power. (8.29 × 10−1 ) × 102 = 8.29 × 101 Concept Problem Revisited The distance between the sun and Neptune would be written as 4.5 × 109 km and the diameter of an electron would be written as 2.2 × 10−13 in. Vocabulary Scientific Notation Scientific notation is a way of writing numbers in the form of a number between 1 and 10 multiplied by a power of 10. The number 196.5 written in scientific notation is 1.965 × 102 and the number 0.0760 written in scientific notation is 7.60 × 10−2 . 346 www.ck12.org Chapter 4. Exponents Guided Practice 1. Express the following product in scientific notation: (4 × 1012 )(9.2 × 107 ) 2. Express the following quotient in scientific notation: 6,400,000 0.008 3. If a = 0.000415, b = 521, and c = 71, 640, find an approximate value for notation. ab c . Express the answer in scientific Answers: 1. Apply the rule (A × 10m ) × (B × 10n ) = (A × B) × (10m+n ) (4 × 1012 ) × (9.2 × 107 ) = (4 × 9.2) × (1012+7 ) (4 × 9.2) × (1012+7 ) = 36.8 × 1019 Express the answer in scientific notation. 36.8 × 1019 = (3.68 × 101 ) × 1019 (3.68 × 101 ) × 1019 = 3.68 × 1020 (4 × 1012 )(9.2 × 107 ) = 3.68 × 1020 2. Begin by expressing the numerator and the denominator in scientific notation. 6.4×106 8.0×10−3 Apply the rule (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m+n ) . (6.4 × 106 ) ÷ (8.0 × 10−3 ) = (6.4 ÷ 8.0) × (106−−3 ) Apply the law of exponents − subtract the exponents of the powers. (6.4 ÷ 8.0) × (106−−3 ) = (0.8) × (109 ) (0.8) × (109 ) = 0.8 × 109 Express the answer in scientific notation. 0.8 × 109 = (8.0 × 10−1 ) × 109 0.8 × 109 = 8.0 × 10−1 × 109 Apply the law of exponents − add the exponents of the powers. 8.0 × 10−1 × 109 = 8.0 × 108 6, 400, 000 = 8.0 × 108 0.008 Express the answer in scientific notation. 3. Express all values in scientific notation. 0.000415 = 4.15 × 10−4 521 = 5.21 × 102 71, 640 = 7.1640 × 104 347 4.6. Scientific Notation www.ck12.org Use the values in scientific notation to determine an approximate value for ab c = ab c . (4.15×10−4 )(5.21×102 ) 7.1640×104 In the numerator, apply the rule (A × 10m ) × (B × 10n ) = (A × B) × (10m+n ) (4.15 × 10−4 )(5.21 × 102 ) (4.15 × 5.21) × (10−4 × 102 ) = 7.1640 × 104 7.1640 × 104 −4 2 (4.15 × 5.21) × (10 × 10 ) 21.6215 × 10−2 = 7.1640 × 104 7.1640 × 104 Apply the rule (A × 10m ) ÷ (B × 10n ) = (A ÷ B) × (10m−n ) . 21.6215 × 10−2 = (21.6215 ÷ 7.1640) × (10−2−4 ) 7.1640 × 104 (21.6215 ÷ 7.1640) × (10−2 × 104 ) = 3.018 × 10−6 Practice Express each of the following in scientific notation: 1. 2. 3. 4. 5. 42,000 0.00087 150.64 56,789 0.00947 Express each of the following in standard form: 6. 7. 8. 9. 10. 4.26 × 105 8 × 104 5.967 × 1010 1.482 × 10−6 7.64 × 10−3 Perform the indicated operations and express the answer in scientific notation 11. 12. 13. 14. 8.9 × 104 + 4.3 × 105 8.7 × 10−4 − 6.5 × 10−5 (5.3 × 106 ) × (7.9 × 105 ) (3.9 × 108 ) ÷ (2.8 × 106 ) For the given values, perform the indicated operations for standard form. 15. . 348 ab c and express the answer in scientific notation and www.ck12.org Chapter 4. Exponents a = 76.1 b = 818, 000, 000 c = 0.000016 16. . a = 9.13 × 109 b = 5.45 × 10−23 c = 1.62 Summary You learned that in an expression like 2x , the "2" is the base and the "x" is the exponent. You learned the following laws of exponents that helped you to simplify expressions with exponents: • • • • • • • • am × an = am+n am m−n (if m > n, a 6= 0) an = a m n (a ) = amn n n n (ab) = aan b a n = bn (b 6= 0) b a0 = 1 (a 6= 0) a−m = a1m √ √ m m n a n = am = n a You learned that scientific notation is a way to express large or small numbers in the form x = a × 10n where 1 ≤ a < 10 and n ∈ Z. You learned that to solve exponential equations with variables in the exponent you should try to rewrite the equations so the bases are the same. Then, set the exponents equal to each other and solve. If the equation has a variable in the base you can try to get rid of the exponent or, make the exponents on each side of the equation the same and then set the bases equal to each other and solve. Finally, you learned all about exponential functions. You learned that for exponential functions of the form y = x ab c + d, if 0 < b < 1 then the function is decreasing and represents exponential decay. If b > 1 then the function is increasing and represents exponential growth. Exponential functions are used in many real-life situations such as with the decay of radioactive isotopes and with interest that compounds. 349 www.ck12.org C HAPTER 5 Polynomials Chapter Outline 5.1 A DDITION AND S UBTRACTION OF P OLYNOMIALS 5.2 M ULTIPLICATION OF P OLYNOMIALS 5.3 S PECIAL P RODUCTS OF P OLYNOMIALS Introduction Here you’ll learn all about polynomials. You’ll start by learning how to add, subtract, and multiply polynomials. Then you will learn how to factor polynomials, which can be thought of as the opposite of multiplying. Next you’ll learn how to divide polynomials and how this connects to factoring. Finally, you’ll learn how to use your graphing calculator to graph polynomials in order to determine the factors and roots of polynomials. 350 www.ck12.org Chapter 5. Polynomials 5.1 Addition and Subtraction of Polynomials Here you’ll learn how to add and subtract polynomials. You are going to build a rectangular garden in your back yard. The garden is 2 m more than 1.5 times as long as it is wide. Write an expression to show the area of the garden. Watch This Khan Academy Adding and Subtracting Polynomials 1 MEDIA Click image to the left for more content. Guidance The word polynomial comes from the Greek word poly meaning “many”. Polynomials are made up of one or more terms and each term must have an exponent that is 0 or a whole number. This means that 3x2 +2x+1 is a polynomial, but 3x0.5 + 2x−2 + 1 is not a polynomial. Some common polynomials have special names based on how many terms they have: • A monomial is a polynomial with just one term. Examples of monomials are 3x, 2x2 and 7. • A binomial is a polynomial with two terms. Examples of binomials are 2x + 1, 3x2 − 5x and x − 5. • A trinomial is a polynomial with three terms. An example of a trinomial is 2x2 + 3x − 4. To add and subtract polynomials you will go through two steps. 1. Use the distributive property to remove parentheses. Remember that when there is no number in front of the parentheses, it is like there is a 1 in front of the parentheses. Pay attention to whether or not the sign in front of the parentheses is + or −, because this will tell you if the number you need to distribute is +1 or −1. 2. Combine similar terms. This means, combine the x2 terms with the x2 terms, the x terms with the x terms, etc. 351 5.1. Addition and Subtraction of Polynomials www.ck12.org Example A Find the sum: (3x2 + 2x − 7) + (5x2 − 3x + 3). Solution: First you want to remove the parentheses. Because this is an addition problem, it is like there is a +1 in front of each set of parentheses. When you distribute a +1, none of the terms will change. 1(3x2 + 2x − 7) + 1(5x2 − 3x + 3) = 3x2 + 2x − 7 + 5x2 − 3x + 3 Next, combine the similar terms. Sometimes it can help to first reorder the expression to put the similar terms next to one another. Remember to keep the signs with the correct terms. For example, in this problem the 7 is negative and the 3x is negative. 3x2 + 2x − 7 + 5x2 − 3x + 3 = 3x2 + 5x2 + 2x − 3x − 7 + 3 = 8x2 − x − 4 This is your final answer. Example B Find the difference: (5x2 + 8x + 6) − (4x2 + 5x + 4). Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a −1 in front of the second set of parentheses. When you distribute a −1, each term inside that set of parentheses will change its sign. 1(5x2 + 8x + 6) − 1(4x2 + 5x + 4) = 5x2 + 8x + 6 − 4x2 − 5x − 4 Next, combine the similar terms. Remember to keep the signs with the correct terms. 5x2 + 8x + 6 − 4x2 − 5x − 4 = 5x2 − 4x2 + 8x − 5x + 6 − 4 = x2 + 3x + 2 This is your final answer. Example C Find the difference: (3x3 + 6x2 − 7x + 5) − (4x2 + 3x − 8) Solution: First you want to remove the parentheses. Because this is a subtraction problem, it is like there is a −1 in front of the second set of parentheses. When you distribute a −1, each term inside that set of parentheses will change its sign.. 1(3x3 + 6x2 − 7x + 5) − 1(4x2 + 3x − 8) = 3x3 + 6x2 − 7x + 5 − 4x2 − 3x + 8 Next, combine the similar terms. Remember to keep the signs with the correct terms. 3x3 + 6x2 − 7x + 5 − 4x2 − 3x + 8 = 3x3 + 6x2 − 4x2 − 7x − 3x + 5 + 8 = 3x3 + 2x2 − 10x + 13 This is your final answer. 352 www.ck12.org Chapter 5. Polynomials Concept Problem Revisited Remember that the area of a rectangle is length times width. Area = l × w Area = (1.5x + 2)x Area = 1.5x2 + 2x Vocabulary Binomial A binomial has two terms that are added or subtracted from each other. Each of the terms of a binomial is a variable (x), a product of a number and a variable (4x), or the product of multiple variables with or without a number (4x2 y + 3). One of the terms in the binomial can be a number. Monomial A monomial can be a number or a variable (like x) or can be the product of a number and a variable (like 3x or 3x2 ). A monomial has only one term. Polynomial A polynomial, by definition, is also a monomial or the sum of a number of monomials. So 3x2 can be considered a polynomial, 2x + 3 can be considered a polynomial, and 2x2 + 3x − 4 can be considered a polynomial. Trinomial A trinomial has three terms (4x2 + 3x − 7). The terms of a trinomial can be a variable (x), a product of a number and a variable (3x), or the product of multiple variables with or without a number (4x2 ). One of the terms in the trinomial can be a number (−7). Variable A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is often referred to as the literal coefficient. Guided Practice 1. Find the sum: (2x2 + 4x + 3) + (x2 − 3x − 2). 353 5.1. Addition and Subtraction of Polynomials 2. Find the difference: (5x2 − 9x + 7) − (3x2 − 5x + 6). 3. Find the sum: (8x3 + 5x2 − 4x + 2) + (4x3 + 7x − 5). Answers: 1. (2x2 + 4x + 3) + (x2 − 3x − 2) = 2x2 + 4x + 3 + x2 − 3x − 2 = 3x2 + x + 1 2. (5x2 − 9x + 7) − (3x2 − 5x + 6) = 5x2 − 9x + 7 − 3x2 + 5x − 6 = 2x2 − 4x + 1 3. (8x3 + 5x2 − 4x + 2) + (4x3 + 7x − 5) = 8x3 + 5x2 − 4x + 2 + 4x3 + 7x − 5 = 12x3 + 5x2 + 3x − 3 Practice For each problem, find the sum or difference. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 354 (x2 + 4x + 5) + (2x2 + 3x + 7) (2r2 + 6r + 7) − (3r2 + 5r + 8) (3t 2 − 2t + 4) + (2t 2 + 5t − 3) (4s2 − 2s − 3) − (5s2 + 7s − 6) (5y2 + 7y − 3) + (−2y2 − 5y + 6) (6x2 + 36x + 13) − (4x2 + 13x + 33) (12a2 + 13a + 7) + (9a2 + 15a + 8) (9y2 − 17y − 12) + (5y2 + 12y + 4) (11b2 + 7b − 12) − (15b2 − 19b − 21) (25x2 + 17x − 23) − (−14x3 − 14x − 11) (−3y2 + 10y − 5) − (5y2 + 5y + 8) (−7x2 − 5x + 11) + (5x2 + 4x − 9) (9a3 − 2a2 + 7) + (3a2 + 8a − 4) (3x2 − 2x + 4) − (x2 + x − 6) (4s3 + 4s2 − 5s − 2) − (−2s2 − 5s + 6) www.ck12.org www.ck12.org Chapter 5. Polynomials 5.2 Multiplication of Polynomials Here you will learn how to multiply polynomials using the distributive property. Jack was asked to frame a picture. He was told that the width of the frame was to be 5 inches longer than the glass width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass and finds the height to width ratio is 4:3. Write the expression to determine the area of the picture frame. Watch This Khan Academy Multiplying Polynomials MEDIA Click image to the left for more content. Guidance To multiply polynomials you will need to use the distributive property. Recall that the distributive property says that if you start with an expression like 3(5x + 2), you can simplify it by multiplying both terms inside the parentheses by 3 to get a final answer of 15x + 6. When multiplying polynomials, you will need to use the distributive property more than once for each problem. Example A Find the product: (x + 6)(x + 5) Solution: To answer this question you will use the distributive property. The distributive property would tell you to multiply x in the first set of parentheses by everything inside the second set of parentheses , then multiply 6 in the first set of parentheses by everything in the second set of parentheses . Here is what that looks like: 355 5.2. Multiplication of Polynomials www.ck12.org Example B Find the product: (2x + 5)(x − 3) Solution: Again, use the distributive property. The distributive property tells you to multiply 2x in the first set of parentheses by everything inside the second set of parentheses , then multiply 5 in the first set of parentheses by everything in the second set of parentheses . Here is what that looks like: 356 www.ck12.org Chapter 5. Polynomials Example C Find the product: (4x + 3)(2x2 + 3x − 5) Solution: Even though at first this question may seem different, you can still use the distributive property to find the product. The distributive property tells you to multiply 4x in the first set of parentheses by everything inside the second set of parentheses, then multiply 3 in the first set of parentheses by everything in the second set of parentheses. Here is what that looks like: Concept Problem Revisited Jack was asked to frame a picture. He was told that the width of the frame was to be 5 inches longer than the glass width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass and finds the height to width ratio is 4:3. Write the expression to determine the area of the picture frame. What is known? • The width is 5 inches longer than the glass • The height is 7 inches longer than the glass • The glass has a height to width ratio of 4:3 The equations: 357 5.2. Multiplication of Polynomials www.ck12.org • The height of the picture frame is 4x + 7 • The width of the picture frame is 3x + 5 The formula: Area = w × h Area = (3x + 5)(4x + 7) Area = 12x2 + 21x + 20x + 35 Area = 12x2 + 41x + 35 Vocabulary Distributive Property The distributive property states that the product of a number and a sum is equal to the sum of the individual 2 products of the number and the addends. For example, in the expression: 3 (x + 5), the distributive property 2 states that the product of a number 3 and a sum (x + 5) is equal to the sum of the individual products of the 2 number 3 and the addends (x and 5). Like Terms Like terms refers to terms in which the degrees match and the variables match. For example 3x and 4x are like terms. Like terms are also known as similar terms. Guided Practice 1. Find the product: (x + 3)(x − 6) 2. Find the product: (2x + 5)(3x2 − 2x − 7) 3. An average football field has the dimensions of 160 ft by 360 ft. If the expressions to find these dimensions were (3x + 7) and (7x + 3), what value of x would give the dimensions of the football field? Answers: 1. (x + 3)(x − 6) 358 www.ck12.org Chapter 5. Polynomials 2. (2x + 5)(3x2 − 2x − 7) 3. Area = l × w 359 5.2. Multiplication of Polynomials www.ck12.org Area = 360 × 160 (7x + 3) = 360 7x = 360 − 3 7x = 357 x = 51 (3x + 7) = 160 3x = 160 − 7 3x = 153 x = 51 The value of x that satisfies these expressions is 51. Practice Use the distributive property to find the product of each of the following polynomials: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 360 (x + 4)(x + 6) (x + 3)(x + 5) (x + 7)(x − 8) (x − 9)(x − 5) (x − 4)(x − 7) (x + 3)(x2 + x + 5) (x + 7)(x2 − 3x + 6) (2x + 5)(x2 − 8x + 3) (2x − 3)(3x2 + 7x + 6) (5x − 4)(4x2 − 8x + 5) 9a2 (6a3 + 3a + 7) −4s2 (3s3 + 7s2 + 11) (x + 5)(5x3 + 2x2 + 3x + 9) (t − 3)(6t 3 + 11t 2 + 22) (2g − 5)(3g3 + 9g2 + 7g + 12) www.ck12.org Chapter 5. Polynomials 5.3 Special Products of Polynomials Here you will learn about special cases of binomial multiplication. A flower is homozygous blue (RR) and another flower is homozygous white (rr). Use a Punnett square to show that a mixture of the two can produce a white flower. Watch This Khan Academy Special Products of Binomials MEDIA Click image to the left for more content. Guidance There are two special cases of multiplying binomials. If you can learn to recognize them, you can multiply these binomials more quickly. Here are the two special products that you should learn to recognize: Special Case 1 (Binomial Squared): (x ± y)2 = x2 ± 2xy + y2 • Example: (x + 5)2 = x2 + 10x + 25 361 5.3. Special Products of Polynomials www.ck12.org • Example: (2x − 8)2 = 4x2 − 32x + 64 Special Case 2 (Difference of Perfect Squares): (x + y)(x − y) = x2 − y2 • Example: (5x + 10)(5x − 10) = 25x2 − 100 • Example: (2x − 4)(2x + 4) = 4x2 − 16 Keep in mind that you can always use the distributive property to do the multiplications if you don’t notice that the problem is a special case. Example A Find the product: (x + 11)2 Solution: This is an example of Special Case 1. You can use that pattern to quickly multiply. (x + 11)2 = x2 + 2 · x · 11 + 112 = x2 + 22x + 121 You can verify that this is the correct answer by using the distributive property: Example B Find the product: (x − 7)2 Solution: This is another example of Special Case 1. You can use that pattern to quickly multiply. 362 www.ck12.org Chapter 5. Polynomials (x − 7)2 = x2 − 2 · x · 7 + 72 = x2 − 14x + 49 You can verify that this is the correct answer by using the distributive property: Example C Find the product: (x + 9)(x − 9) Solution: This is an example of Special Case 2. You can use that pattern to quickly multiply. (x + 9)(x − 9) = x2 − 92 = x2 − 81 You can verify that this is the correct answer by using the distributive property: 363 5.3. Special Products of Polynomials www.ck12.org Concept Problem Revisited Each flower will have one-half of the blue genes and one-half of the white genes. Therefore the equation formed will be: 0.5B + 0.5W The offspring will have the genetic makeup (the mixture produced) using the equation: (0.5B + 0.5W )2 364 www.ck12.org Chapter 5. Polynomials Notice that this is an example of Special Case 1. You can expand the offspring genetic makeup equation to find out the percentage of offspring (or flowers) that will be blue, white, or light blue. Therefore 25% of the offspring flowers will be blue, 50% will be light blue, and 25% will be white. Vocabulary Distributive Property The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: 3(x + 5), the distributive property states that the product of a number (3) and a sum (x + 5) is equal to the sum of the individual products of the number (3) and the addends (x and 5). Guided Practice 1. Expand the following binomial: (x + 4)2 . 2. Expand the following binomial: (5x − 3)2 . 3. Determine whether or not each of the following is a difference of two perfect squares: a) a2 − 16 b) 9b2 − 49 c) c2 − 60 Answers: 1. (x + 4)2 = x2 + 4x + 4x + 16 = x2 + 8x + 16. 2. (5x − 3)2 = 25x2 − 15x − 15x + 9 = 25x2 − 30x + 9 3. a)Yes, a2 − 16 = (a + 4)(a − 4) b) Yes, 9b2 − 49 = (3b + 7)(3b − 7) c) No, 60 is not a perfect square. 365 5.3. Special Products of Polynomials www.ck12.org Practice Expand the following binomials: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. (t + 12)2 (w + 15)2 (2e + 7)2 (3z + 2)2 (7m + 6)2 (g − 6)2 (d − 15)2 (4x − 3)2 (2p − 5)2 (6t − 7)2 Find the product of the following binomials: 11. 12. 13. 14. 15. (x + 13)(x − 13) (x + 6)(x − 6) (2x + 5)(2x − 5) (3x + 4)(3x − 4) (6x + 7)(6x − 7) Summary You learned that adding, subtracting, and multiplying polynomials all rely on the distributive property. You also learned that factoring is the reverse of multiplying because when you factor a polynomial you are trying to rewrite the polynomial as a product of other polynomials. You learned how to factor completely by first looking for common factors and then using other methods to factor the remaining expression. You learned special cases of factoring to watch out for including the difference of perfect squares, perfect square trinomials, and the sum and difference of cubes. You learned how factoring can allow you to solve a quadratic equation with the help of the zero product property. You learned how to divide polynomials and how dividing polynomials can connect to factoring. You learned about the factor theorem and that polynomial long division can actually help you to factor higher degree polynomials. Finally, you learned that factors of polynomials connect to their x-intercepts, and that you can use a graphing calculator to graph polynomial functions in order to find out more information about the factors of the polynomial. 366